Question

помогите решить, даю все свои баллы

Answer 1

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Answer 2

Ответ:

1) $-\frac{1}{4(x-2)^2}$

2) $\frac{x+y}{a+y}$

3) x = $\frac{19}{13}$ или $1\frac{6}{13}$ .

Объяснение:

1) $\frac{4}{4-x^2} * \frac{x+2}{16(x-2)}$ = $\frac{1}{(2-x)(2+x)} * \frac{x+2}{4(x-2)}$ = $\frac{1}{2-x} * \frac{1}{4(x-2)}$ = $\frac{1}{4(2-x)(x-2)}$ = $\frac{1}{4*(-(x-2))(x-2)}$ = $\frac{1}{-4(x-2)(x-2)}$ = $\frac{1}{-4(x-2)^2}$ .

2) $\frac{x^2-y^2+ax+ay}{a^2+xy+ax-y^2}$ = $\frac{(x-y)(x+y)+a(x+y)}{(a-y)(a+y)+x(y+a)}$ = $\frac{(x+y)(x-y+a)}{(a+y)(a-y+x)}$ = $\frac{x+y}{a+y}$ .

3)

$\frac{5x-2}{3} - \frac{4x-3}{5} = \frac{6}{5}$

5(5x - 2) - 3(4x - 3) = 18

25x - 10 - 12x + 9 = 18

13x - 1 = 18

13x = 18 + 1

13x = 19

x = $\frac{19}{13}$ или $1\frac{6}{13}$ .